A vibration magnetometer placed in magnetic meridian has a small bar magnet

A Vibration Magnetometer Placed In Magnetic Meridian Has A Small Physics Question

A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth’s horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be

Options

(a) 1 s
(b) 2 s
(c) 3 s
(d) 4 s

Correct Answer:

4 s

Explanation:

Time period of a vibration magnetometer,
T 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁ √(B₁ / B₂)
2 √ (24 x 10⁻⁶ / 6 x 10⁻⁶) = 4 sec.

Related Questions:

  1. An alternating emf given by equation e=300 sin⁡[(100 π)t] V
  2. A Gaussian surface in the cylinder of cross-section πa² and length L is immersed
  3. When a mass M is attached to a spring of force constant K,
  4. An A.C source is rated at 220 V, 50 Hz. The time taken for voltage to change
  5. Bragg’s law for X-rays is

Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*